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wariber [46]
3 years ago
13

Another model for a growth function for a limited population is given by the Gompertz function, which is a solution to the diffe

rential equation dPdt=cln(KP)P where c is a constant and K is the carrying capacity. Answer the following questions. 1. Solve the differential equation with a constant c=0.05, carrying capacity K=3000, and initial population P0=1000.
Mathematics
1 answer:
krek1111 [17]3 years ago
6 0

Answer:

The solution is

P(t) = 1000e^{-1}e^{e^{-0.05t}}.

Step-by-step explanation:

We have the Gompertz's differential equation

\frac{dP}{dt} = -cP\ln(KP).

Notice that this a separable equation. So,

\frac{dP}{P\ln(KP)} = -cdt.

Now, integrating in both sides:

\int\frac{dP}{P\ln(KP)} = -\int cdt.

Recall that

\int\frac{dP}{P\ln(KP)} = \ln(|\ln(KP)|).

Thus,

\ln(|\ln(KP)|) = -ct+A,

and taking exponential:

|\ln(KP)| = A'e^{-ct}.

Taking another exponential

P(t) = \frac{A''}{K} e^{e^{-ct}}.

Now, let us substitute the given values for c and K:

P(t) = \frac{A''}{3000} e^{e^{-0.05t}}.

To find the value of the constant A'' we use the initial condition:

P(0) = \frac{A''}{3000} e^{e^{0}} = \frac{A''}{3000}e=1000.

Then,

A'' = \frac{3\cdot 10^6}{e}.

Hence, the solution is

P(t) = 1000e^{-1}e^{e^{-0.05t}}

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